# Tag Info

## New answers tagged collision

1

Assuming that hands remain completely static, and the object do not breaks, then all of its energy can be considered to be converted into heat and sound (as you have already described).

-1

Energy is accepted by molecules of both objects and turns into heat form.

2

You're making a mistake in assuming that there is any left. Heat and sound account for all of it. There is one exception, though. If it crashes into something, and that breaks or bends the object, then the potential energy of the molecules is higher. That's why cars end up smashed after a collision. The molecules of the metal or plastic or whatever have ...

1

In answer to the main question, matter does, in fact, "pass" through other matter. Starting from the macro scale (stars , galaxies), down to the micro scale (atoms), it happens all the time. The "free" movement of matter starts to get impeded, as the atoms start making latices (solids, crystals). But even at this scale, as Rutherford demonstrated, matter ...

1

If I throw a small rock (m = 1kg) at a big rock (100kg) the small rock rebounds. Let's say my weight is 80kg, if I would jump into a big rock instead of bouncing back I would move in the same direction as a big rock. The big rock is heavier but it is not reflecting me. Why is that? There are two conditions which are to be met if a body A ...

1

The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...

1

First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...

1

Two laws govern collisions: conservation of momentum and conservation of energy. Momentum is the product of mass and velocity, so we can write $$\sum m_i\cdot \vec{v_i} = const$$ Conservation or energy is a little bit trickier, since energy can be converted from one type to another. In an elastic collision, the kinetic energy is conserved, so $$\sum ... 1 You can be sure that at least one reasearch team may help answer your question. Try to mail them. It seems that one can make such model. But still, as we can see it's a bit probabilistic field. http://trb.metapress.com/content/v4t5712601175275/ ... 1 Consider the following results: From the definition of scalar product of four vectors,$$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$The usual dispersion relations:$$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$The velocity \textbf{v}_i in terms of momentum and energy:$$ \tag{3} ...

1

Assuming a direct hit - so traveling through about 50 km of atmosphere - at 0.1 c that would take about 2 ms if it didn't get slowed down too much by the atmosphere. What about drag force? Let's assume a radius $r$, density $\rho$, mass $m = \frac43 \pi r^3 \rho$. If it is a sphere, it experiences a drag force $F=\frac12 \rho_a v^2 C_d A$. Putting $\rho_a=1 ... 0 Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum ... 0 Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected. 1 The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on ... 2 *elastic collision occur between atomic particles? inelastic collision occur between ordinary objects? perfectly inelastic collision occur during shooting? super elastic collision occur during explosion?* as John Rehnnie has explained, an elementary particle as the term implies, is not made of other particles, has no lattices ... 0 Initial velocity is provided by an external force acting on the object to move it from rest. Their initial velocity can be the same, however since the mass is different their acceleration will be different. -1 In equillibrium state the initial velocity of all bodies is equal .bcoz of both the bullets are at rest so their initial velocities are also equal whether both have same or different mass. 2 The only factor is the capacity to return to the original shape when it is deformed by an external force. An elastic object recovers its shape after it has been compressed and deformed, if you crush a plastic bottle and remove the cap it will partially return to his original length, a lead ball is almost completely irreversibly deformed. The Coefficient ... 1 This follows on from my answer to your previous question: Factors on which Coefficient of restitution depend. The coefficient of restitution of a collision depends on the available degrees of freedom for energy to be lost. If you take your example of the collision of atomic particles, let's say two electrons, then there isn't anywhere for the initial ... 0 A purely elastic collision is a theoretical construct, which helps to simplify some situations. In all collisions, some energy is lost as heat, and thus the collision is said to be inelastic. Perfectly elastic collision does not occur between atomic particles Your problem here is that you are assuming atomic collisions can be represented by classical ... 0 Most of us will have discovered that a glass rings if you tap it, and the sound of the ring will persist for several seconds after hitting the glass. This shows that elastic energy stored in the glass is not dissipated fast. I would guess that fewer people have tried the experiment with a lead beaker, but I have and I can report that the lead does not ring. ... 2 Conservation of momentum works here like everywhere else. When A (with mass$m_A$) throws a ball with mass$m_b$with velocity$v$, then$v_A=-v_b\frac{m_b}{m_A}$so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards$B$at velocity$v$but instead at$v-v_A$since$A$started moving backwards... When B ... 1 Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ... 2 A lot of the answers get distracted by the "feeling pain" part of the question. So let me start by focusing on that - simplifying a very complex psycho-physiological issue to a simple physical statement (this is a physics forum - we leave the other stuff for other sites): During the impact of two bodies, there will be an exchange of momentum. The force ... 2 What is the difference that leads to conservation of kinetic energy in elastic collision ? The difference is only in the properties of the material of a body. If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored ... 2 The simple answer is that in an elastic collision (for objects >> in mass than typical molecules) energy moves from kinetic to potential then back to kinetic as long as the "elastic limits" of the materials are not exceeded. In other words, as long as they act like springs. In non-elastic collision the energy goes mostly from kinetic of the colliding masses ... 8 Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa. 4 No mater the inertial referential: The pain is linked to the energy dissipated by the change of speed of the 2 objects in any inertial referential. Remarque 1: a part of the energy can be absorbed by the ball by deformation or heating. So a soft ball would make less pain. Remarque 2: The pain depends on what is static behind you and what is static behind ... 29 Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ... -1 Yes you will.. what really matters here is the relative momentum ! 2 Pavel Petrman gave a different perspective, but since impact remains same pain should remain same, given that properties of the situation remain same. To try take a ball in your hand and run towards a weighing scale and hit it and in next case exchange ball and weighing scale and repeat the procedure. Take a note of readings on weighing scale. 13 Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames. However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if ... 1 Although I think the question is related more to algorithm (as a branch of programming), but since it incorporates aspects of physics, I will give it a go. OP, know that there is not one answer to your problem. It can be accomplished in several ways. Here I would introduce you to some important variables in your scenario: The kinetic energy of the body in ... -1 impulse is what you feel and force is what you receive. A whip feels more than a punch. If punch is delivered with the same speed as whip and then the hand is withdrawn back quickly then there is no doubt punch will be more effective.. but withdrawing the hand instantaneously is difficult practically, by minimizing the time of contact, practically speaking ... 2 I have heard that we can maximize the force by minimizing the time of impact in a punch That is not true. if you want to crack a bone hit at the right place with right timing.... but if you want to break bone the other way.. have a good mass and bang on with full speed, no need to pull the hand backwards like karate chop... I got the ... 1 Just to clarify, impulse is force times time. Which one matters depends on what you are trying to achieve. If you want to propell a ball thru the air, then it's the impulse that matters. Note that a ball whacked for a short time by a bat at high force can go further than you can throw by applying lower force and much longer time. If you are trying to ... 1 Let's say that the ball strikes the floor with the linear momentum$-p_1$(you see I take the direction of the momentum in consideration). Some part of the momentum is lost (some energy is lost to the floor) s.t. the ball rises again, but at the floor level it has a linear momentum$+p_2$. So, what the floor "gets" in this scenario? It GETS a linear momentum ... 3 If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one ( B ),from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball ( A )? In order to get a clear picture, let's consider the extreme case when the velocity of B = 0 Let's ... 1 by conservation of linear momentum the center of mass of the system will keep moving, thus the two ball are together they will keep moving in the same direction than the faster ball, at the same speed than the center of mass, that is, slower than the initial speed of the faster ball. Assuming a totally elastic collision, after the balls push back each other, ... 2 Think of it this way: Ball A is moving at 10 m/s Ball B is moving at 3 m/s Both balls have the same mass Here we have Ball A collides with Ball B, transferring energy. During this transfer (ignore the deformation for now, since that doesn't seem to be an issue at the moment, so we've got an elastic collision), consider what happens in the exchange of ... 0 Since the block are on a frictionless surface, they don't interact with the Earth, so it doesn't make much sense to talk about the "block-earth" system. Within the context of this block-block system, the mechanical energy is always conserved. In terms of equations, at the point of collision we have$m_{1}v_{10}=m_{1}v_{1}+m_{2}v_{2}$... 0 First of all, I agree with the result you obtained using conversation of energy. Your logic with the Work-Energy Theorem does not work because the total work done on block A is not$-1/2kx_0^2$(I am denoting the max compression by$x_0$). Suppose that block A first comes into contact with the spring at$s=0$and suppose that its velocity reaches$5.5\, ...

Top 50 recent answers are included